I am working on the following question. It involves finding a proof for a trig identity using linear algebra. The proof is one involving $sin(\alpha +\theta)$ and $cos(\alpha +\theta)$, as you will see. I will go through where I am up to, progressing through each part of the question.
Let $T_{\alpha}$ be the linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$
which is the rotation counterclockwiseby $\alpha$, and $T_{\theta}$ the counterclockwise rotation by θ.(A) Write down the standard matrices for $T_{\alpha}$ and $T_{\theta}$, explain your reasoning.
Let the standard matrix for $T_{\alpha}$ be $A$ and let the standard matrix for $T_{\theta}$ be $B$, then
$$A=\begin{bmatrix}
\cos (\alpha) & -\sin (\alpha) \\
\sin (\alpha) & \cos (\alpha) \\
\end{bmatrix}$$
and
$$B=\begin{bmatrix}
\cos (\theta) & -\sin (\theta) \\
\sin (\theta) & \cos (\theta) \\
\end{bmatrix}$$
On to the next part of the queston.
(B) Explain what the linear transformation $T_{\alpha} \circ T_{\theta}$ does to $\mathbb{R}^2$.
It first rotates a given point by $\alpha$ degrees and then rotates the given point by $\theta$ degrees.
(C) Compute the matrix for $T_{\alpha} \circ T_{\theta}$ by multiplying the matrices for $T_{\alpha}$ and $T_{\theta}$
So,
$$AB = \begin{bmatrix}
\cos (\alpha)\cos (\theta) – \sin (\alpha) \sin (\theta) & -\cos (\alpha)\sin (\theta) – \sin (\alpha) \cos (\theta) \\
\sin (\alpha)\cos (\theta) + \cos (\alpha) \cos (\theta) & -\sin (\alpha)\sin (\theta) + \cos (\alpha) \cos (\theta) \\
\end{bmatrix}$$
(D) On the other hand, from the description in part (b), you can directly write down
the matrix for $T_{\alpha} \circ T_{\theta}$. What is that matrix?
If that matrix is $C$, then
$$C=\begin{bmatrix}
\cos (\alpha + \theta) & -\sin (\alpha + \theta) \\
\sin (\alpha + \theta) & \cos (\alpha + \theta) \\
\end{bmatrix}$$
(E) Since the matrices from parts (c) and (d) are describe the same linear transformation, they must be equal. What identities among sin and cos must therefore be
true?
So I must set $AB=C$. Then
$$\begin{bmatrix}
\cos (\alpha)\cos (\theta) – \sin (\alpha) \sin (\theta) & -\cos (\alpha)\sin (\theta) – \sin (\alpha) \cos (\theta) \\
\sin (\alpha)\cos (\theta) + \cos (\alpha) \cos (\theta) & -\sin (\alpha)\sin (\theta) + \cos (\alpha) \cos (\theta) \\
\end{bmatrix}=\begin{bmatrix}
\cos (\alpha + \theta) & -\sin (\alpha + \theta) \\
\sin (\alpha + \theta) & \cos (\alpha + \theta) \\
\end{bmatrix}$$
These are the identities I was looking for! Now the next part has me worried.
Using a similar idea, find formulas for $\sin(3\theta)$ and $\cos(3\theta)$ in terms of $\sin(\theta)$ and $\cos(\theta)$.
Now I am not completely hopeless – I was able to come up with this next bit. But I am not sure if I have done things correctly.
I'll use the transformation $T_{\theta}$ from before. The standard matrix is
$$A=\begin{bmatrix}
\cos (\alpha) & -\sin (\alpha) \\
\sin (\alpha) & \cos (\alpha) \\
\end{bmatrix}$$
I thought that maybe if I transformed a point three times it would rotate it $3\theta$ degrees, i.e with a standard matrix $AAA$. The result was
$$AAA=B$$
$$=$$$$ \begin{bmatrix}
\cos^{3}(\theta)-\cos(\theta)\sin^2(\theta)-2\cos(\theta)\sin^2(\theta) & -\sin(\theta)\cos^2(\theta)+\sin^3(\theta)-2\cos^2(\theta)\sin(\theta) \\
2\sin(\theta)\cos^2(\theta)-\sin^3(\theta)+\cos^2(\theta)\sin(\theta) & -2\sin^3(\theta)\cos(\theta)-\sin^2(\theta)cos(\theta)+\cos^3(\theta)\ \\
\end{bmatrix}$$
As before I would say that $AAA$ is equal to a transformation matrix
$$A'=\begin{bmatrix}
\cos (3\alpha) & -\sin (3\alpha) \\
\sin (3\alpha) & \cos (3\alpha) \\
\end{bmatrix}$$
Then setting $A'=B$ would result in some identities. They just seem quite long and I am not sure if what I have done is correct or checks out. Any help would be appreciated.
Best Answer
Thanks for explaining your reasoning in that much detail. Let me give some remarks